Problem: $1+\dfrac{1}{\sqrt[ 3]2}+\dfrac{1}{\sqrt[ 3]3}+\dfrac{1}{\sqrt[ 3]4}+...+\dfrac{1}{\sqrt[ 3]n}+...$ Is the series convergent or divergent? Choose 1 answer: Choose 1 answer: (Choice A) A Convergent (Choice B) B Divergent
This is a $p$ -series, because it's of the general form $\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{^p}}$. Specifically, this is a $p$ -series where $p=\dfrac{1}{3}$. $p$ -series are convergent for $p>1$ and divergent for $0<p\leq1$. [What about non-positive p-values?] In conclusion, the series is divergent.